In: Chemistry
Which of the following atoms would you predict to have a fourth-ionization energy very much greater than the third?
Na |
Mg |
Al |
Si |
P |
A:- Element Atomic number Outer most electronic configuration
Na 11 3s1
Mg 12 3s2
Al 13 3s2 3p1
Si 14 3s2 3p2
P 15 3s2 3p3
First three ionnisation energy for the removal of three electrons from the Al metal atom to form Al^3+ ion, it has noble gas, Ne configuration (stable elctronic configuration). i.e. 1s2 2s2 2p6
Al -----------> Al^3+ + 3e
The fourth electron remove from the noble gas electronic configuration, so it required very much greater energy than the third.
Al^3+ electronic configuaration is equal [Ne] configuration
A:- Al